Krull–Akizuki theorem

In algebra, the Krull–Akizuki theorem states the following: let A be a at most one-dimensional reduced noetherian ring,[1] K its total ring of fractions. If B is a subring of a finite extension L of K containing A then B is a one-dimensional noetherian ring. Furthermore, for every nonzero ideal I of B, ${\displaystyle B/I}$ is finite over A.[2]

Note that the theorem does not say that B is finite over A. The theorem does not extend to higher dimension. One important consequence of the theorem is that the integral closure of a Dedekind domain A in a finite extension of the field of fractions of A is again a Dedekind domain. This consequence does generalize to a higher dimension: the Mori–Nagata theorem states that the integral closure of a noetherian domain is a Krull domain.

Proof

Here, we give a proof when ${\displaystyle L=K}$. Let ${\displaystyle {\mathfrak {p}}_{i}}$ be minimal prime ideals of A; there are finitely many of them. Let ${\displaystyle K_{i}}$ be the field of fractions of ${\displaystyle A/{{\mathfrak {p}}_{i}}}$ and ${\displaystyle I_{i}}$ the kernel of the natural map ${\displaystyle B\to K\to K_{i}}$. Then we have:

${\displaystyle A/{{\mathfrak {p}}_{i}}\subset B/{I_{i}}\subset K_{i}}$.

Now, if the theorem holds when A is a domain, then this implies that B is a one-dimensional noetherian domain since each ${\displaystyle B/{I_{i}}}$ is and since ${\displaystyle B=\prod B/{I_{i}}}$. Hence, we reduced the proof to the case A is a domain. Let ${\displaystyle 0\neq I\subset B}$ be an ideal and let a be a nonzero element in the nonzero ideal ${\displaystyle I\cap A}$. Set ${\displaystyle I_{n}=a^{n}B\cap A+aA}$. Since ${\displaystyle A/aA}$ is a zero-dim noetherian ring; thus, artinian, there is an l such that ${\displaystyle I_{n}=I_{l}}$ for all ${\displaystyle n\geq l}$. We claim

${\displaystyle a^{l}B\subset a^{l+1}B+A.}$

Since it suffices to establish the inclusion locally, we may assume A is a local ring with the maximal ideal ${\displaystyle {\mathfrak {m}}}$. Let x be a nonzero element in B. Then, since A is noetherian, there is an n such that ${\displaystyle {\mathfrak {m}}^{n+1}\subset x^{-1}A}$ and so ${\displaystyle a^{n+1}x\in a^{n+1}B\cap A\subset I_{n+2}}$. Thus,

${\displaystyle a^{n}x\in a^{n+1}B\cap A+A.}$

Now, assume n is a minimum integer such that ${\displaystyle n\geq l}$ and the last inclusion holds. If ${\displaystyle n>l}$, then we easily see that ${\displaystyle a^{n}x\in I_{n+1}}$. But then the above inclusion holds for ${\displaystyle n-1}$, contradiction. Hence, we have ${\displaystyle n=l}$ and this establishes the claim. It now follows:

${\displaystyle B/{aB}\simeq a^{l}B/a^{l+1}B\subset (a^{l+1}B+A)/a^{l+1}B\simeq A/{a^{l+1}B\cap A}.}$

Hence, ${\displaystyle B/{aB}}$ has finite length as A-module. In particular, the image of I there is finitely generated and so I is finitely generated. Finally, the above shows that ${\displaystyle B/{aB}}$ has zero dimension and so B has dimension one. ${\displaystyle \square }$

References

1. In this article, a ring is commutative and has unity.
2. Bourbaki 1989, Ch VII, §2, no. 5, Proposition 5 harvnb error: no target: CITEREFBourbaki1989 (help)

• Nicolas Bourbaki, Commutative algebra